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Quantitative Aptitude Prep Tips for MBA

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3 months ago

Quantitative Aptitude Prep Tips for MBA Class 12th

If P₁ + P₂ + P₃ = 59, where P₁, P₂ and P₃ are prime numbers and P₁ < P₂ < P₃, then how many values can P₃ take?

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Payal Gupta

Contributor-Level 10

P₁ + P₂ + P₃ = 59

P₁ and P₂ → minimize

Then P₃ → minimum i.e. 47

P₁	P₂	P₃
5	7	47
5	11	43
3	13	43
5	13	41
5	17	37
3	19	37
5	23	31
7	23	29

So, only possible value of P₃ can take → 29, 31, 37, 41, 43, and 47

Total 6 values

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3 months ago

Quantitative Aptitude Prep Tips for MBA Class 12th

How many pairs of positive integers x, y satisfy $\frac{1}{x} + \frac{6}{y} = \frac{1}{15}$ , where y is a two digit odd integer?

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Payal Gupta

Contributor-Level 10

$\frac{1}{x} = \frac{1}{15} - \frac{6}{y}$

$\frac{1}{x} = \frac{y - 90}{15 y}$

$x = \frac{15 y}{y - 90}$ ______equation 1

i.e. y = 90+y,

1≤ y₁ ≤9

y₁ is odd

Equation 1 becomes

$x = \frac{15 (90 + y_{1})}{y_{1}}$

$= 15 + \frac{90 * 15}{y_{1}}$

y₁ can take 1,3,5,9

y=91, 93, 95, 99

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3 months ago

Quantitative Aptitude Prep Tips for MBA Class 12th

An alcohol solution is formed by mixing two solutions. Solution 1 had a concentration of 20 percent and a volume of 3L while solution 2 had a concentration of 40 percent with a volume of 1L. The resultant solution was named solution 3. Three litres of solution 4 which had an unknown concentration was mixed with solution 3. What must be concentration of solution 4 if that the new resultant solution has a concentration of 50 percent?

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Payal Gupta

Contributor-Level 10

Sol1	20percent (0.6)	3L
Sol2	40percent (0.4)	1L
Sol3	1L	4L
Sol4	XL	3L
Sol5	3.5L	7L

1 + x = 3.5

x= 2.5

Required percent y = $\frac{2.5}{3} * 100$ = 83.33 percent

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3 months ago

Quantitative Aptitude Prep Tips for MBA Class 12th

Two geometric means a, b are inserted between x and y such that a³+ b³ = n (ab) A,where A is the arithmetic mean of x and y. What is value of n?

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Payal Gupta

Contributor-Level 10

x, a, b, y are in GP

a = xr

b = xr²

y = xr³

r = ${(\frac{y}{x})}^{1 / 3}$

rewrite,

a = x^2/3.y^1/3

b = x^1/3.y^2/3

a * b = x * y

a³+ b³= x²y + y² x

= xy (x+y)

= xy (2A) …………… $[\frac{x + y}{2} = A]$

Hence, n = 2

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3 months ago

Quantitative Aptitude Prep Tips for MBA Class 12th

A square with maximum possible length of the side is inscribed in an equilateral triangle and a circle of maximum possible radius is inscribed in the square. Find the ratio of the length of equilateral of the equilateral triangle to the radius of circle.

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Payal Gupta

Contributor-Level 10

Let the side of = a

CD = a/2

Let EC = x

In GEC

GE/EC = tan60 = $\sqrt[]{3}$

GE = x $\sqrt[]{3}$

DE = a/2 – x = FG = HF

GE = HG

$x \sqrt[]{3} = \frac{a}{2} - x + \frac{a}{2} - x$

x $\sqrt[]{3}$ = a – 2x

x = $\frac{a}{\sqrt[]{3} + 2}$

GE = x $\sqrt[]{3}$ = $\frac{a \sqrt[]{3}}{\sqrt[]{3} + 2}$

r = side of square ÷ 2

r= $\frac{a \sqrt[]{3}}{4 + 2 \sqrt[]{3}}$

ratio = a/r

= $\frac{a (4 + 2 \sqrt[]{3})}{a \sqrt[]{3}}$

= $\frac{4 + 2 \sqrt[]{3})}{\sqrt[]{3}} = \frac{4}{\sqrt[]{3}} + 2$

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3 months ago

Quantitative Aptitude Prep Tips for MBA Class 12th

A certain sum of money invested at CI becomes Rs 3380 in 2 years & Rs 3515.20 in 3 years. What is the sum invested & what is rate % per annum?

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Vishal Baghel

Contributor-Level 10

(d):3515.2 – 3380 = 135.2 = SI on Rs. 3380 for one year.

By SI formula, 135.2 = $3380 * R * \frac{1}{100}$ $\frac{}{}$ or R = 4 %

Now we have 3380 = P (1.04)2 or P = Rs. 3125

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3 months ago

Quantitative Aptitude Prep Tips for MBA Class 12th

Duronto Express in motion, 66 metre long overtakes a passenger train 88 metres long travelling at 30 kmph in the opposite direction in 0.168 minute. Find speed of the Duronto Express.

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Vishal Baghel

Contributor-Level 10

(c):Let speed of passenger train be x m/s.

$[(\frac{66 + 88}{8.33 + x})] = 0.168 * 60$

$[30 * \frac{5}{18} = 8.33]$

x = 6.94 m/sec = 25 kmph.

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3 months ago

Quantitative Aptitude Prep Tips for MBA Class 12th

A 10 floor building is build by 2 men, 3 women and 4 boys in 10 days. If a man, a women & a boy work in the ratio 5 : 4 : 2. Find the time that 6 men, 4 women and 7 boys take to build a 16 floor building.

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Vishal Baghel

Contributor-Level 10

(c):We have 4 men = 5 women 1 man = $\frac{5}{4}$ $\frac{}{}$ women, 2 women = 4 boys, 1 women = 2 boys $\frac{5}{4}$ $\frac{}{}$ women = 2 * $\frac{5}{4}$ $\frac{}{}$ boys = $\frac{5}{2}$ $\frac{}{}$ or 1 man = $\frac{5}{4}$ $\frac{}{}$ women = $\frac{3}{2}$ $\frac{}{}$ boys

Now 2M + 3W+ 4B = 2 * B + 3 * 2B + 4B = 15 B Boys

Or 15 Boys do the work in 10 days (10 hectares) 6M + 4W + 7B = (6 * + 4 * 2 + 7) B = 30

Boys 30 boys will do 16 hectares of work in 8 days.

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3 months ago

Quantitative Aptitude Prep Tips for MBA Class 12th

During the months of October Krishna went to a temple every day. He offered 4 flowers to her favourite deity. At the end of the month, he noticed that he did not offer the same combination of flowers on any two days during the month. What is the minimum number of varieties of flowers offered by Krishna during the month?

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Payal Gupta

Contributor-Level 10

Let he offered 'n' different flowers

October has 31 days

ⁿC₄ ³ 31

Minimum value of n can take 7

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